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3 years ago

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An insulated rigid tank is divided into two compartments by a partition. One compartment contains 4 kg of O2 at 20 ºC and 150 kP

a, and the other compartment contains 7 kg of N2 at 40 ºCand 100 kPa. Now the partition is removed, and two gases are allowed to mix.
Determine: (a) the mixture temperature, (b) number of moles of each gas, (c) initial volume occupied by each gas, and (d) final pressure of the mixture. Take the values of specific heat of both gases at 300 K.

2 answers:

valkas [14]3 years ago

5 0

Answer:

Explanation:

Given that,

Mass of oxygen M = 4kg

Temperature of oxygen T.i = 20°C

T.i = 20+273 =293K

Specific heat capacity of oxygen

C = 0.658 J/kgK

And at pressure P = 150Kpa

Mass of Nitrogen m = 7kg

Temperature of Nitrogen t.i = 40°C

t.i = 40+273 = 313K

And at pressure p = 100Kpa

Specific heat capacity of nitrogen

c = 0.743 J/kgK

a. Mixture temperature can be determine from the energy balance of the container

Using Heat energy formula

∆U = 0

For oxygen

H.o = MC∆θ

H.o = MC(T.f - T.i)

H.o = 4×0.658 (T.f - 293)

H.o = 2.632(T.f - 293)

For Nitrogen

H.n = mc∆θ

H.n = mc(T.f - t.i)

H.n = 7×0.743(T.f-313)

H.n = 5.201(T.f-313)

Then,

∆U = 0

H.o + H.n = 0

2.632(T.f - 293) + 5.201(T.f-313) = 0

2.632T.f -771.176+ 5.201T.f-1627.9=0

7.833T.f - 2399.089 = 0

7.833T.f = 2399.089

T.f = 2399.089/7.833

T.f = 306.28 K

b. Number of molecules n?

Number of mole can be calculated by using the formula

n = mass/molar mass

For oxygen,

Molar mass of oxygen O2

O2 = 16×2 = 32g/mol

n = mass of oxygen / molar mass of oxygen

n = 4000/32

n = 125 mole

For Nitrogen

Molar mass of nitrogen N.2

N.2 = 14×2 = 28g/mol

n = mass of nitrogen /molar mass of nitrogen

n' = 7000/28

n' = 250 mole

C. The initial volume of each gas V?

Volume can be calculated using the ideal gas law

PV = nRT

V = nRT/P

R is a constant = 8.314 J/mol•K

For nitrogen

V = n'Rt/p

V = 250 × 8.314 ×313 / 100,000

Vn = 6.51m³

For Oxygen

V = nRT/P

V = 125 × 8.314 × 293 / 150,000

Vo = 2.03m³

d. Final pressure mixture

Molar mass of mixture

M = total mass/total mole

M = (7000+4000)/(250+125)

M = 29.333g/mol

The mixture total number of mole

N = mole of oxygen + mole of nitrogen

N = 250+125 = 375 mol

Then, using ideal gas law

PV = nRT

P = nRT/V

Total mole = 375mol

R = 8.314 J/mol•K

Mixture Temperature T.f = 306.28K

Total Volume Vo+Vn = 2.03+6.51

V = 8.54m³

Then,

P = nRT/V

P = 375×8.314×306.28/8.54

P = 111,815.5 Pa

To K•pa

P = 111.82 K•Pa

o-na [289]3 years ago

4 0

Answer:

Find answers attached

Explanation:

Assuming both gases are idea gas

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Answer:

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Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

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integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

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so, 3 = 0 - 0 + C

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C = -4

So,

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A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is given by

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

So,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

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$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

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Where P is the number of poles of the turbo-generator.

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4 0

3 years ago

Write equations used to calculate the diode reverse saturation current, the voltage at which diode goes into resistive behavior,

laiz [17]

Answer:

Diode equation for reverse saturation current

$I_o = A\times e^{\frac{-Eg}{KT}} + B\times e^{\frac{-Eg}{2KT}}$

Voltage at which diode goes into Resistive region:V=-5 volts

Voltage at which high level injection occurs:Va=0.55 volt

Voltage at which avalanche multiplication occurs:V=5volts

Explanation:

we take here forward and reverse 0.7 volt and -5 volt

As Diode current equation is express as

$I_D = I_o \times (e^{\frac{V_D}{\eta V_T}} -1 )$ ....................1

here $I_D$ is total current through the diode and $I_o$ is reverse saturated current and $V_D$ is voltage drop across diode and $\eta$ is idealized factor and $V_T$ is thermal voltage

so here we know that when Bios is forward than

$V_D$ = $V_T$ .................2

ans Bios is Reverse than

$V_D$ = $V_R$ ..................3

so here

1. diode reverse saturation current is express as

$I_o = A\times e^{\frac{-Eg}{KT}} + B\times e^{\frac{-Eg}{2KT}}$

and

2. Voltage at which diode go into Reverse behavior will be

$V_D$ = $V_R$ = -5 volt

and

3. voltage at which high level injection occur that is

Va = 0.55 volt

and

4. voltage at which avalanche multiplication occurs is

Va = 5 volt

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